# Limits and continuity

## Homework Statement

find limit of
x1/3y2 / x + y3
as x,y tends to 0,0

## The Attempt at a Solution

i realise i can't use limits of individual variable since the denominator goes to 0 if x,y goes to 0,0

i realise i can't use squeeze theorem since the demnominator is not square, so negative numbers come into play

i realise that if i do a substitution of z = x1/3
i get
zy2 / z3 +y3
which seems to be what the question is hinting... but i get stuck... anyone can help? thanks!

## Homework Statement

the next problem is to find all points that are continuous in the function f

f(x,y) = (y-5)cos(1/x2) if x not = 0
if x = 0, then f(x,y) = 0

## The Attempt at a Solution

my notes says that to show continuity, i must show that f(x,y) = f(a,b) when x,y tends to a,b
how do i do that?
does it mean i do something like this

-1< cos(1/x2) <1
(y-5) < (y-5)cos( 1/x2) < (y-5)
so all points are continuous except at x = 0 ?

BUt for f(x,y) = 0 when x = 0, it is not continous right? since cos(1/0) = undefined?

so the function is continous at all points except x= 0?

thanks!

hunt_mat
Homework Helper
Suppose you let $$x^{\frac{1}{3}}=y=t$$ and then examined the limit as t tents to zero, what do you get then?

Suppose you let $$x^{\frac{1}{3}}=y=t$$ and then examined the limit as t tents to zero, what do you get then?

i get t t2 / t3 +t3 = 1/2
so they tend to 1/2?

but how can i let t be = x1/3 and also t = y at the same time?

Char. Limit
Gold Member
i get t t2 / t3 +t3 = 1/2
so they tend to 1/2?

but how can i let t be = x1/3 and also t = y at the same time?

It's one of the paths you can follow. Now try going along the path x=0.

It's one of the paths you can follow. Now try going along the path x=0.

if i use x = 0 i get 0 / y3 ? which is 0?
if i use y = 0 i get 0 / x = 0

er i still don't udnerstand what you mean by x1/3 = y = t
i thought the method of going from x=0 and then y tends to something is to prove the function has no limits?

oh and any idea for the 2nd question? thanks!

Char. Limit
Gold Member
if i use x = 0 i get 0 / y3 ? which is 0?
if i use y = 0 i get 0 / x = 0

er i still don't udnerstand what you mean by x1/3 = y = t
i thought the method of going from x=0 and then y tends to something is to prove the function has no limits?

Basically, you can follow along multiple paths for a limit. The t is just a dummy variable to indicate that you're following the path y=x1/3. This path and the path x=0 have different limit values, so...

hunt_mat
Homework Helper
What char. Limit said.

Oh!

so i am moving along the line x1/3 = y for the first case to get limit of 1/2

and when x = 0 , i get limit of 0

so since they have different limits, then limit doesn't exist?

thanks!

btw, is there a definite way of "seeing" the route that will have a different limit to x = 0 because i only can see simple stuff like x= 0 and y= 0... or does it all boils down to practice and experience?

Char. Limit
Gold Member
You've got it perfectly!